Question

If $a_r$ is the coefficient of $x^{10-r}$ in the Binomial expansion of $(1+x)^{10}$, then $\displaystyle\sum_{r=1}^{10} r^3\left(\frac{a_r}{a_{r-1}}\right)^2$ is equal to

• A. 3025
• B. 1210
• C. 5445
• D. 4895

Answer 1

The Correct Option is B

To solve this problem, we start by understanding the binomial expansion of \((1+x)^{10}\). The general term of this expansion is given by:

\(T_{r+1} = \binom{10}{r} x^r = a_r x^r\)

Here, \(a_r\) represents the coefficient of \(x^r\), which is \(\binom{10}{r}\).

In this question, we are specifically interested in the coefficient of \(x^{10-r}\). Hence \(a_r\) corresponds to the coefficient \(\binom{10}{10-r}\).

The task here is to evaluate:

\(\sum_{r=1}^{10} r^3 \left(\frac{a_r}{a_{r-1}}\right)^2\)

First, evaluate the expression \(\frac{a_r}{a_{r-1}}\):

\(\frac{a_r}{a_{r-1}} = \frac{\binom{10}{10-r}}{\binom{10}{10-(r-1)}} = \frac{\binom{10}{r}}{\binom{10}{r-1}} = \frac{10-r+1}{r}\)

Thus,

\(\left(\frac{a_r}{a_{r-1}}\right)^2 = \left(\frac{10-r+1}{r}\right)^2 = \left(\frac{11-r}{r}\right)^2\)

The desired expression becomes:

\(\sum_{r=1}^{10} r^3 \left(\frac{11-r}{r}\right)^2 = \sum_{r=1}^{10} r^3 \frac{(11-r)^2}{r^2} = \sum_{r=1}^{10} \frac{r^3 (11-r)^2}{r^2} = \sum_{r=1}^{10} r(11-r)^2\)

Now calculate this sum:

Expand \((11-r)^2\):

\((11-r)^2 = 121 - 22r + r^2\)

The expression for each term becomes:

\(r(11-r)^2 = r(121 - 22r + r^2) = 121r - 22r^2 + r^3\)

So we evaluate:

\(\sum_{r=1}^{10} (121r - 22r^2 + r^3)\)

The calculation simplifies to:

\(\sum_{r=1}^{10} 121r = 121 \sum_{r=1}^{10} r = 121 \cdot 55 = 6655\)

\(\sum_{r=1}^{10} 22r^2 = 22 \cdot 385 = 8470\) (since \(\sum_{r=1}^{10} r^2 = 385\))

\(\sum_{r=1}^{10} r^3 = 3025\) (since \(\sum_{r=1}^{10} r^3 = (55)^2\))

Combine these results:

\(6655 - 8470 + 3025 = 1210\)

Therefore, the final answer to the problem is 1210.